Projectile Motion - Skier & Ramp

1. The problem statement, all variables and given/known data
A ski jumper acquires a speed of 119.0 km/hr by racing down a steep hill. He then lifts off into the air from a horizontal ramp. Beyond this ramp, the ground slopes downward at an angle of θ = 45 degrees.

2. Relevant equations
Assuming the skier is in free-fall motion after he leaves the ramp, at what distance d down the slope does the skier land?

3. The attempt at a solution
I converted 119.0 km/hr to m/s, so 33.055 m/s

I then try to find the horizontal range: 0=33.055^2 + 2*9.8*X and found X=55.74m
55.74cos45=78.83m down the ramp.

With respect to the slope which is angled at 45 dgree, the ski jumper is jumping off the ramp with a velocity u=(119x5/18)m/s at angle of 45 degree!
Thus, the horizontal range he covers on the slope, d=(u^2xsin(2θ)/g).
Correct me if im wrong,or you can ask for further doubts or queries.

With respect to the slope which is angled at 45 dgree, the ski jumper is jumping off the ramp with a velocity u=(119x5/18)m/s at angle of 45 degree!
Thus, the horizontal range he covers on the slope, d=(u^2xsin(2θ)/g).
Correct me if im wrong,or you can ask for further doubts or queries.

You are wrong. Horizontal means horizontal - it doesn't mean parallel to the slope.
Gravity continues to act vertically, not orthogonally to the slope.

You are wrong. Horizontal means horizontal - it doesn't mean parallel to the slope.
Gravity continues to act vertically, not orthogonally to the slope.

Thanks for the glitch.
Ok, suppose 'd' is the hypotenuse of an isosceles triangle of angle 45 degree and its equal sides=dcos45 or (dsin45 whatever). Then time taken for the skii jumper to cover 'd' is t=(2xh/g)½ = (2xdcos45/g)½.
Now, R=dsin45=u x t= (u^2 x 2 x dcos45/g)½
solving out the above equation will give the value of 'd'.